HDU 4392 反素数java

网上找的模板能求10^80次方以内的。

import java.io.BufferedInputStream; import java.math.BigInteger; import java.util.ArrayList; import java.util.HashMap; import java.util.LinkedList; import java.util.List; import java.util.Map; import java.util.Queue; import java.util.Scanner;  class Node {     private static final int MAXP = 60;      public BigInteger K;     public long F;     public int N;     public int[] A;      public Node()     {         K = BigInteger.ZERO;         A = new int[MAXP];     } }  public class Main{     private static final int MAXIP = 250;     private static final int MAXP = 60;      private static BigInteger[] prime;      private static void init()     {         boolean[] isPrime = new boolean[MAXIP];         for(int i=0;i<MAXIP;++i)         {             isPrime[i] = true;         }         isPrime[0] = isPrime[1] = false;         for(int i=4;i<MAXIP;i+=2)         {             isPrime[i] = false;         }         for(int i=3;i<MAXIP;i+=2)         {             for(int j=3;i*j<MAXIP;j+=2)             {                 isPrime[i*j] = false;             }         }         prime = new BigInteger[MAXP];         for(int i=0, j=0;i<MAXIP;++i)         {             if(isPrime[i])             {                 prime[j++] = BigInteger.valueOf(i);             }         }     }      public static void main(String args[])     {         init();         List<BigInteger> P = new ArrayList<BigInteger>();         BigInteger MP = BigInteger.ZERO;         List<Node> ans = new ArrayList<Node>();         Scanner cin = new Scanner(new BufferedInputStream(System.in));         while(cin.hasNext())         {             BigInteger temp = cin.nextBigInteger();             P.add(temp);             if(temp.compareTo(MP) == 1)             {                 MP = temp;             }             ans.add(new Node());         }         Map<Long, BigInteger> map = new HashMap<Long, BigInteger>();         Queue<Node> queue = new LinkedList<Node>();         Node origin = new Node();         origin.K = BigInteger.ONE;         origin.F = 1;         origin.N = 0;         queue.add(origin);         map.put(origin.F, origin.K);         while(!queue.isEmpty())         {             Node u = queue.peek();             queue.remove();             BigInteger compare = map.get(u.F);             if(compare != null)             {                 if(compare.compareTo(u.K) == -1)                 {                     continue;                 }             }             for(int i=0;i<P.size();++i)             {                 if(u.K.compareTo(P.get(i)) <= 0)                 {                     if(u.F > ans.get(i).F)                     {                         ans.get(i).F = u.F;                         ans.get(i).K = u.K;                     }                     else if(u.F == ans.get(i).F)                     {                         if(u.K.compareTo(ans.get(i).K) == -1)                         {                             ans.get(i).K = u.K;                         }                     }                 }             }             for(int i=0;i<u.N;++i)             {                 Node v = new Node();                 v.K = u.K.multiply(prime[i]);                 if(v.K.compareTo(MP) <= 0)                 {                     v.F = u.F / (u.A[i] + 1) * (u.A[i] + 2);                     v.N = u.N;                     for(int j=0;j<u.N;++j)                     {                         v.A[j] = u.A[j];                     }                     ++ v.A[i];                     boolean flag = true;                     compare = map.get(v.F);                     if(compare != null)                     {                         if(compare.compareTo(v.K) <= 0)                         {                             flag = false;                         }                         else                         {                             map.remove(v.F);                         }                     }                     if(flag)                     {                         queue.add(v);                         map.put(v.F, v.K);                     }                 }             }             Node v = new Node();             v.K = u.K.multiply(prime[u.N]);             if(v.K.compareTo(MP) <= 0)             {                 v.F = u.F * 2;                 v.N = u.N + 1;                 for(int i=0;i<u.N;++i)                 {                     v.A[i] = u.A[i];                 }                 ++ v.A[u.N];                 boolean flag = true;                 compare = map.get(v.F);                 if(compare != null)                 {                     if(compare.compareTo(v.K) <= 0)                     {                         flag = false;                     }                     else                     {                         map.remove(v.F);                     }                 }                 if(flag)                 {                     queue.add(v);                     map.put(v.F, v.K);                 }             }         }         for(int i=0;i<ans.size();++i)         {             System.out.println(ans.get(i).K.toString() + " " + ans.get(i).F); //第一个数为满足因子个数最多的K，第二个数为K的因子个数        }     } }